So I'm doing this problem from Stats class.
Let $S$ be the side of a regular hexagon. $$S \sim \text{Uniform}(10,14)$$
I have to compute the mean and pdf of the Area.
I've looked it up, and the area of a regular hexagon of side $S$ is given by $$A=\frac{3\sqrt3}{2}s^2$$
Therefore they're asking me to compute the pdf and mean of A.
So $$S = +\sqrt{\frac{2}{3\sqrt{3}}A}$$
The pdf of $S$ is given by $$f_S(x) = \frac{1}{14-10} = 0.25 $$
The pdf of A would be given by
$$f_A(y) = f_S(S) \ \cdot |S'| =0.25 \cdot \frac{\sqrt{2}\sqrt{3}}{6\sqrt{y\sqrt{3}}}$$
Then the mean is given by:
$$\int_{150\sqrt{2}}^{294 \sqrt{3}} y f_A(y) = 434.3576510$$.
But I've done a simulation with RStudio, and the value I obtain is around $378.4282$
$$f_A(a) = f_S(g^{-1}(a)) \left|\frac{dg^{-1}}{da}\right|$$ with $a = g(s) = c s^2$ where $c = 3 \sqrt{3}/2$. Thus $$f_A(a) = \frac{f_S(\sqrt{a/c})}{2\sqrt{ca}},$$ and since $$f_S(s) = \frac{1}{4} \mathbb 1 (10 \le s \le 14),$$ the desired PDF of $A$ is $$f_A(a) = \frac{1}{8 \sqrt{ca}} \mathbb 1 (100c \le a \le 196c) = \frac{\mathbb 1 (150 \sqrt{3} \le c \le 294 \sqrt{3})}{4 \cdot 2^{1/2} \cdot 3^{3/4} a^{1/2}}.$$ The mean is $$\operatorname{E}[A] = 218 \sqrt{3}.$$ Your error was using the lower bound $150\sqrt{2}$ instead of $150\sqrt{3}$.
Alternatively, you could have calculated the mean by noting $$\operatorname{E}[A] = \operatorname{E}[c S^2] = c \operatorname{E}[S^2] = c \left(\operatorname{Var}[S] + \operatorname{E}[S]^2\right),$$ and recalling that the mean and variance for a uniform distribution are $$\operatorname{E}[S] = \frac{a+b}{2} = 12, \quad \operatorname{Var}[S] = \frac{(b-a)^2}{12} = \frac{4}{3}.$$